Maclaurin 2018 Q5
5. For which integers \(n\) is \(\dfrac{16\,\bigl(n^2 - n - 1\bigr)^2}{2n - 1}\) also an integer?
Maclaurin Olympiad 2018, Question 5. © UK Mathematics Trust. Original paper on ukmt.org.uk
Answer: \(n = -12\), \(-2\), 0, 1, 3 and 13
We set \(m = 2n - 1\), so \(n = \tfrac{1}{2}(m + 1)\). Now the expression becomes
\(\begin{aligned}\frac{16}{m}\left[\left(\frac{m + 1}{2}\right)^2 - \left(\frac{m + 1}{2}\right) - 1\right]^2 &= \frac{1}{m}\bigl[(m + 1)^2 - 2\,(m + 1) - 4\bigr]^2\\[6pt] &= \frac{1}{m}\bigl[m^2 - 5\bigr]^2 = \frac{m^4 - 10m^2 + 25}{m}.\end{aligned}\)
This is an integer if, and only if, \(m\) is a factor of 25. Hence the values of \(m\) are \(\pm 1\), \(\pm 5\), \(\pm 25\) and the corresponding values of \(n\) are \(-12\), \(-2\), 0, 1, 3 and 13.
Instead of using the substitution, the numerator can be rewritten as \(\bigl((2n - 1)^2 - 5\bigr)^2\) to obtain the same result.
Solution © UK Mathematics Trust, from the Maclaurin Olympiad 2018 solutions.